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Equilibrium - Homogeneous Equilibria

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Homogeneous equilibria are physical or chemical equilibria in which all the reacting substances and products are present in the same phase (e.g., all gases or all in a single liquid solution).

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For a general reversible reaction aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD, the equilibrium constant in terms of molar concentration is denoted as KcK_c.

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For reactions involving gases, the equilibrium constant can be expressed in terms of partial pressures, denoted as KpK_p.

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The relationship between KpK_p and KcK_c is given by Kp=Kc(RT)ΔngK_p = K_c(RT)^{\Delta n_g}, where Δng\Delta n_g is the difference between the number of moles of gaseous products and gaseous reactants.

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The value of the equilibrium constant is independent of initial concentrations but depends on the temperature of the system.

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If Δng=0\Delta n_g = 0, then Kp=KcK_p = K_c (e.g., in the synthesis of HIHI from H2H_2 and I2I_2).

📐Formulae

Kc=[C]c[D]d[A]a[B]bK_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}

Kp=(PC)c(PD)d(PA)a(PB)bK_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}

Kp=Kc(RT)ΔngK_p = K_c(RT)^{\Delta n_g}

Δng=(nc+nd)−(na+nb) (only for gaseous components)\Delta n_g = (n_c + n_d) - (n_a + n_b) \text{ (only for gaseous components)}

Pi=χi×Ptotal (Partial pressure using mole fraction)P_i = \chi_i \times P_{\text{total}} \text{ (Partial pressure using mole fraction)}

💡Examples

Problem 1:

For the reaction: 2NOCl(g)⇌2NO(g)+Cl2(g)2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g) at 500 K500\text{ K}, the value of KcK_c is 3.75×10−63.75 \times 10^{-6}. Calculate KpK_p for the reaction at the same temperature. (Given R=0.0821 L atm K−1mol−1R = 0.0821\text{ L atm K}^{-1}\text{mol}^{-1})

Solution:

  1. Identify Δng\Delta n_g: Δng=moles of gaseous products−moles of gaseous reactants=(2+1)−2=1\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = (2 + 1) - 2 = 1.
  2. Use the relation: Kp=Kc(RT)ΔngK_p = K_c(RT)^{\Delta n_g}.
  3. Substitute values: Kp=(3.75×10−6)×(0.0821×500)1K_p = (3.75 \times 10^{-6}) \times (0.0821 \times 500)^1.
  4. Calculation: Kp=3.75×10−6×41.05=1.539×10−4K_p = 3.75 \times 10^{-6} \times 41.05 = 1.539 \times 10^{-4}.

Explanation:

Since the reaction involves gases and there is an increase in the number of moles (Δng>0\Delta n_g > 0), KpK_p will be numerically greater than KcK_c when RT>1RT > 1.

Problem 2:

Calculate the equilibrium constant KcK_c for the reaction H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g) if at equilibrium [H2]=0.10 mol L−1[H_2] = 0.10\text{ mol L}^{-1}, [I2]=0.20 mol L−1[I_2] = 0.20\text{ mol L}^{-1}, and [HI]=0.40 mol L−1[HI] = 0.40\text{ mol L}^{-1}.

Solution:

  1. Write the KcK_c expression: Kc=[HI]2[H2][I2]K_c = \frac{[HI]^2}{[H_2][I_2]}.
  2. Substitute the equilibrium concentrations: Kc=(0.40)2(0.10)(0.20)K_c = \frac{(0.40)^2}{(0.10)(0.20)}.
  3. Solve: Kc=0.160.02=8K_c = \frac{0.16}{0.02} = 8.

Explanation:

In this reaction, Δng=2−(1+1)=0\Delta n_g = 2 - (1+1) = 0. Therefore, the units of concentration cancel out, making KcK_c dimensionless in this specific case, and Kp=KcK_p = K_c.